 # Calculating pH of Aqueous Salt Solutions Chemistry Tutorial

## Key Concepts

• pH of an aqueous solution of a salt of a strong monoprotic acid and strong base is 7 (at 25°C)(1)

Cation does not undergo hydrolysis (react with water).

Anion does not undergo hydrolysis (react with water).

• pH of an aqueous solution of a salt of a weak monoprotic acid(2) and strong base is >7 (at 25°C)

⚛ Cation does not undergo hydrolysis (react with water).

⚛ Anion does undergo hydrolysis (react with water).
Hydroxide ions (OH-(aq)) are a product of the hydrolysis reaction:

A-(aq) + H2O(l) ⇋ HA(aq) + OH-(aq)

Steps to calculate the pH of an aqueous solution of the salt (MA) of a weak acid (HA) and a strong base (MOH):

Step 1: Calculate the initial concentration in mol L-1 of the salt, [MA], and hence of the anion, [A-].

Step 2: Write the equation for the hydrolysis of the anion.

Step 3: Use a R.I.C.E. Table to determine the equilibrium concentration of all species.

Step 4: Calculate the value of the anion hydrolysis constant (Kh) (base dissociation constant, Kb) if not given.

Step 5: Calculate the concentration of hydroxide ions using Kb

Step 6: Calculate the pOH of the solution.

Step 7: Calculate the pH of the solution.

• pH of an aqueous solution of a salt of a strong monoprotic acid and weak base is <7 (at 25°C)

⚛ Anion does not hydrolyse (react with water).

⚛ Cation does undergo hydrolysis (react with water).
Hydrogen ions (H+(aq)) or oxidanium ions(3) (H3O+(aq)) are a product of the hydrolysis reaction.

M+(aq) + H2O(l) ⇋ MOH(aq) + H+(aq)

Steps to calculate the pH of an aqueous solution of the salt (MA) of a strong acid (HA) and a weak base (MOH):

Step 1: Calculate the initial concentration of the salt, [MA], and hence of the cation, [M+].

Step 2: Write the equation for the hydrolysis of the cation.

Step 3: Use a R.I.C.E. Table to determine the equilibrium concentration of all species.

Step 4: Calculate the value of the cation hydrolysis constant (Kh) (acid dissociation constant, Ka) if not given.

Step 5: Calculate the concentration of hydrogen ions using Ka

Step 6: Calculate the pH of the solution.

• pH of an aqueous solution of a salt of a weak monoprotic acid and weak base depends on the relative strengths of the acid and base since both the cation and the anion will undergo hydrolysis:

⚛ Kh(cation) > Kh(anion) then solution is acidic (pH < 7 at 25°C).

⚛ Kh(cation) = Kh(anion) then solution is neutral (pH = 7 at 25°C).

⚛ Kh(cation) < Kh(anion) then solution is basic (pH > 7 at 25°C).

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## pH of Aqueous Solutions of Salts of Strong Acid and Strong Base

Sodium chloride, NaCl, is the salt of a strong acid (HCl(aq)) and strong base (NaOH(aq)).
It is produced when the strong acid hydrochloric acid (HCl(aq)), reacts with the strong base, sodium hydroxide (NaOH(aq)) in a neutralisation reaction as shown in the balanced chemical equation below:

 general word equation: word equation: balanced chemical equation: acid + base → salt + water hydrochloric acid + sodium hydroxide → sodium chloride + water HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

When all the acid has been neutralised by all the base the resulting solution contains only sodium chloride dissolved in water.
We could evaporate off all the water to produce solid sodium chloride salt.

Imagine we dissolved a small amount of solid sodium chloride, NaCl(s), in water, H2O(l).
Sodium chloride (NaCl(s)) is very soluble in water so the sodium chloride dissociates in water producing sodium ions (Na+(aq)) and chloride ions (Cl-(aq)) according to the following balanced chemical equation:

NaCl(s) → Na+(aq) + Cl-(aq)

The sodium ions do not hydrolyse (do not react with water).
IF sodium ions hydrolysed, the products would be sodium hydroxide (NaOH(aq)) and hydrogen ions (H+(aq)) as shown below:

Na+(aq) + H2O(l) ⇋ NaOH(aq) + H+(aq)

But we know that sodium hydroxide (NaOH(aq)) is a strong base that fully dissociates in water:

NaOH(aq) → Na+(aq) + OH-(aq)

so if Na+ reacted with water, the NaOH(aq) formed would immediately dissociate to re-form the Na+(aq).
We can say with confidence that sodium ions, Na+, DO NOT hydrolyse (do not react with water).

The chloride ions do not hydrolyse (do not react with water).
IF chloride ions hydrolysed, the products would be hydrochloric acid (HCl(aq)) and hydroxide ions (OH-(aq)) according to the following balanced chemical equation:

Cl-(aq) + H2O(l) ⇋ HCl(aq) + OH-(aq)

But we know that hydrochloric acid (HCl(aq)) is a strong acid that fully dissociates in water:

HCl(aq) → H+(aq) + Cl-(aq)

so if Cl- reacted with water, the HCl(aq) formed would immediately dissociate to re-form the Cl-(aq).
We can say with confidence that chloride ions, Cl-, DO NOT hydrolyse (do not react with water).

So when we dissolve the salt of a strong acid and strong base in water the ONLY species contributing to the pH of the resultant solution is the water.
This is because water molecules undergo self-dissociation (self-ionisation or auto-dissociation) according to the following chemical equation:

H2O(l) ⇋ H+(aq) + OH-(aq)

For the self-dissociation of water, the ion product (Kw) for the reaction is:

Kw = [H+(aq)][OH-(aq)]

and since water is neutral, [H+(aq)] =[OH-(aq)]

Kw = [H+(aq)]2

so the concentration of hydrogen ions is:

[H+(aq)] = √Kw

at 25°C, Kw = 10-14, so

[H+(aq)] = √10-14 = 10-7 mol L-1

Since pH = −log10[H+(aq)] then

pH = −log10[10-7] = 7

Which leads to this important general result:

At 25°C, the pH of an aqueous solution of the salt of a monoprotic strong acid and strong base is 7.

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## pH of Aqueous Solutions of Salts of Weak Acid and Strong Base

Sodium acetate (sodium ethanoate), CH3COONa, is the salt of a weak acid, acetic acid (ethanoic acid), CH3COOH(aq), and a strong base, sodium hydroxide, NaOH(aq).
Sodium hydroxide, NaOH(aq), is a strong base because it fully dissociates in water:

NaOH(aq) → Na+(aq) + OH-(aq)

Acetic acid (ethanoic acid), CH3COOH(aq), is a weak acid because it only partially dissociates in water:

CH3COOH(aq) ⇋ CH3COO-(aq) + H+(aq)     Ka = 1.8 × 10-5 (at 25°C)

Sodium acetate is produced when acetic acid reacts with sodium hydroxide in a neutralisation reaction according to the following chemical equation:

 general word equation: word equation: balanced chemical equation: acid + base → salt + water acetic acid (ethanoic acid) + sodium hydroxide → sodium acetate (sodium ethanoate) + water CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)

When all the available acetic acid has been neutralised by all the available sodium hydroxide, the resulting solution is an aqueous solution of sodium acetate.

Because sodium acetate is soluble in water, we could make a similar aqueous solution of sodium acetate by adding a small amount of solid sodium acetate to water.
When we do this, the intial solution will be made up of sodium ions, Na+(aq), and acetate ions, CH3COO-(aq)

We discovered in the section above that the sodium ions, Na+(aq), will not react with water (do not hydrolyse).

Will acetate ions, CH3COO-(aq), react with water?
IF acetate ions react with water then the products of the reaction would be acetic acid, CH3COOH(aq), and hydroxide ions, OH-(aq), as shown in the chemical equation below:

CH3COO-(aq) + H2O(l) ⇋ CH3COOH(aq) + OH-(aq)

Because acetic acid is a weak acid, some (not all!) of the acetic acid produced by the hydrolysis of acetate ions will dissociate, so the concentration of acetate ions, acetic acid and hydroxide ions reach an equilibrium position.
Acetate ions DO react with water! Acetate ions DO hydrolyse!

The expression for the equilibrium constant for the hydrolysis of acetate ions, Kh, (equivalent to the base dissociation of acetate ions, Kb) is given below:

 Kb = [CH3COOH(aq)][OH-(aq)][CH3COO-(aq)] = Kh

Unfortunately we can't easily look up a table of Kb values, but we can calculate the value of Kb at 25°C :

 Kb = [CH3COOH(aq)][OH-(aq)][CH3COO-(aq)] multiply numerator and denominator by [H+(aq)]: Kb = [CH3COOH(aq)][OH-(aq)][CH3COO-(aq)] × [H+(aq)][H+(aq)] Kb = [CH3COOH(aq)][CH3COO-(aq)][H+(aq)] × [OH-(aq)][H+(aq)] The term on the left is the reciprocal of the acetic acid dissociation constant, Ka. The term on the right is the expression for the ion product of water. Kb = 1 Ka × Kw Ka = 1.8 × 10-5 (25°C) Kw = 10-14 (25°C) Kb = 1       1.8 × 10-5 × 10-14 Kb = 5.6 × 10-10

And of course, the water also undergoes self-dissociation to produce hydrogen ions and hydroxide ions!

When we dissolve sodium acetate in water, there are two reversible reactions involved in establishing the pH of the final solution:

1. hydrolysis of acetate ions

CH3COO-(aq) + H2O(l) ⇋ CH3COOH(aq) + OH-(aq)     Kb = 5.6 × 10-10 (at 25°C)

2. self-dissociation of water

H2O(l) ⇋ H+(aq) + OH-(aq)     Kw = 10-14 (at 25°C)

Since Kb is much greater than Kw, the hydrolysis of acetate ions is the major contributor to the concentration of hydroxide ions in solution, that is, we ignore the negligible amount of hydroxide ions contributed by the self-dissociation of water.

In order to determine the pH of an aqueous solution of sodium acetate we will need to know the concentration of acetate ions.

Let's add 1.00 gram of sodium acetate to enough water to make 1.00 L of solution.

Calculate the moles of sodium acetate in 1.00 g

n(CH3COONa) = moles = mass (g) ÷ molar mass (g mol-1

m(CH3COONa) = mass = 1.00 g

M(CH3COONa) = molar mass = 12.01 + (3 × 1.008) + 12.01 + (2 × 16.00) + 22.99 = 82.034 g mol-1

n(CH3COONa) = 1.00 g ÷ 82.034 g mol-1 = 0.0122 mol

Calculate the intial concentration of sodium acetate in solution:

c(CH3COONa) = n(CH3COONa) ÷ V(CH3COONa)

n(CH3COONa) = 0.0122 mol

Vn(CH3COONa) = 1.00 L

c(CH3COONa) = 0.0122 mol ÷ 1.00 L = 0.0122 mol L-1

Calculate the initial concentration of acetate ions in solution:

CH3COONa → CH3COO-(aq) + Na+(aq)

CH3COONa is soluble in water, so [CH3COONa] = [CH3COO-(aq)] = [Na+(aq)] = 0.0122 mol L-1

Use a R.I.C.E. Table to determine the equilibrium concentrations of species in the solution as a result of the hydrolysis (base dissociation) of acetate ions.
Let x be the change in concentration, then the concentration of acetate ions will decrease by x while the concentrations of acetic acid and hydroxide ions will both increase by x:

 Reaction Initial Concentration (mol L-1) Change in Concentration (mol L-1) Equilibrium Concentration (mol L-1) CH3COO-(aq) + H2O(l) ⇋ CH3COOH(aq) + OH-(aq) 0.0122 0 0 −x +x +x 0.0122 −x ≈ 0.0122 x x Assume x << 0.0122, that is, x is negligible compared to 0.0122

Calculate the concentration of hydroxide ions, x

 Kb = [CH3COOH(aq)][OH-(aq)][CH3COO-(aq)] 5.6 × 10-10 = [x][x][0.0122] 5.6 × 10-10 × 0.0122 = x2 6.8 × 10-12 = x2 √6.8 × 10-12 = √x2 2.6 × 10-6 = x

[OH-(aq)] = x = 2.6 × 10-6 mol L-1

Calculate the pOH of the equilibrium solution:

pOH = −log10[OH-(aq)] = −log10[2.6 × 10-6] = 5.6

So now we know the pOH of the solution. But how do we calculate the pH?
Remember we said the contribution to the hydroxide ion concentration by the self-dissociation of water was negligible?
The same cannot be said for the contribution to the hydrogen ion concentration!
And since Kw = [H+][OH-] = 10-14
Then: -log10[H+] + -log10[OH-] = -log10[10-14]
At 25°C, pH + pOH = 14
Therefore ...

Calculate the pH of the equilibrium solution:

For an aqueous solution at 25°C: pH = 14 − pOH = 14 − 5.6 = 8.4

In general there are 7 steps to calculate the pH of an aqueous solution of the salt of a weak acid and a strong base at 25°C:

Step 1: Calculate the initial concentration of the salt, [MA], and hence of the anion, [A-].

Step 2: Write the equation for the hydrolysis of the anion.

Step 3: Use a R.I.C.E. Table to determine the equilibrium concentration of all species.

Step 4: Calculate the value of the anion hyrolysis constant (base dissociation constant, Kb) if not given.

Step 5: Calculate the concentration of hydroxide ions using Kb

Step 6: Calculate the pOH of the solution.

Step 7: Calculate the pH of the solution.

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## pH of Aqueous Solutions of Salts of Strong Acid and Weak Base

Ammonium chloride, NH4Cl, is the salt of a strong acid, hydrochloric acid (HCl(aq)), and a weak base, ammonia (NH3(aq)) .
Hydrochloric acid is a strong acid, it completely dissociates in water: HCl(aq) → H+(aq) + Cl-(aq)
Ammonia (NH3(aq)) is a weak base, it displays some tendency to accept a proton (H+(aq)) to form the ammonium ion (NH4+(aq)):

NH3(aq) + H2O(l) ⇋ NH4+(aq) + OH-(aq)     Kb = 1.8 × 10-5

Ammonium chloride, NH4Cl, can be produced when an aqueous solution of ammonia is neutralised by hydrochloric acid as shown in the chemical equation below:

NH3(aq) + HCl(aq) → NH4Cl(aq)

The resulting solution is an aqueous solution of ammonium chloride, NH4Cl(aq)

We could make a similar aqueous solution of ammonium chloride by dissolving solid ammonium chloride in water. Because ammonium chloride is soluble in water, initially we would have a solution of ammonium ions and chloride ions.

NH4Cl(aq) → NH4+(aq) + Cl-(aq)

We have already learnt above that chloride ions do not react with water, do not hydrolyse.
But ammonium ions will react with water, will hydrolyse because ammonium is the conjugate acid of a weak base.
The chemical equation for the hydrolysis of ammonium ions is given below (equivalent to the acid dissociation of ammonium ions):

NH4+(aq) + H2O(l) ⇋ NH3(aq) + H3O+(aq)

and

 Kh = [NH3(aq)][H3O+(aq)][NH4+(aq)] = Ka

It is unlikely that you will find the value of Ka for this reaction tabulated, but you can calculate it:

Ka(ammonium) = Kw ÷ Kb(ammonia)

At 25°C:

Ka(ammonium) = 10-14 ÷ 1.8 × 10-5 = 5.6 × 10-10

Let's dissolve 1.00 g of solid ammonium chloride in enough water to make 1.00 L of solution and calculate the pH of the resultant solution.

First we need to calculate the amount of ammonium chloride in moles:

n(NH4Cl) = moles = mass (g) ÷ molar mas (g mol-1)

m(NH4Cl) = mass = 1.00 g

M(NH4Cl) = molar mass = 14.01 + (4 × 1.008) + 35.45 = 53.492 g mol-1

n(NH4Cl) = 1.00 g ÷ 53.492 g mol-1 = 0.0187 mol

Calculate the initial concentration of NH4Cl, and hence of NH4+(aq)

c(NH4Cl) = n(NH4Cl) ÷ V(solution)

n(NH4Cl) = 0.0187 mol

V(solution)= volume in L = 1.00 L

c(NH4Cl) 0.0187 mol ÷ 1.00 L = 0.0187 mol L-1

Since ammonium chloride completely dissociates into ions in water:

NH4Cl → NH4+(aq) + Cl-(aq)

The initial concentration of ammonium ions is equal to the concentration of ammonium chloride:

[NH4+(aq)] = c(NH4Cl) = 0.0187 mol L-1

Use a R.I.C.E. Table to determine the equilibrium concentrations of species in the solution as a result of the hydrolysis (acid dissociation) of ammonium ions.
Let x represent the change in concentration, then the concentration of ammonium ions will decrease by x while the concentration of ammonia and and oxidanium ions will both increase by x.

 Reaction Initial Concentration (mol L-1) Change in Concentration (mol L-1) Equilibrium Concentration (mol L-1) NH4+(aq) + H2O(l) ⇋ NH3 + H3O+(aq) 0.0187 0 0 −x +x +x 0.0187 −x ≈ 0.0187 x x Assume x << 0.0187, that is, x is negligible compared to 0.0187

Calculate the concentration of hydrogen ions (oxidanium ions, H3O+), x

 Ka = [NH3(aq)][H3O+(aq)][NH4+(aq)] 5.6 × 10-10 = [x][x][0.0187] 5.6 × 10-10 × 0.0187 = x2 1.0 × 10-11 = x2 √1.0 × 10-11 = √x2 3.2 × 10-6 = x

[H3O+(aq)] = x = 3.2 × 10-6 mol L-1

Calculate pH of solution: pH = −log10[H3O+(aq)]

pH = −log10[3.2 × 10-6] = 5.5

In general there are 6 steps to calculate the pH of an aqueous solution of the salt of a strong acid and a weak base at 25°C

Step 1: Calculate the initial concentration of the salt, [MA], and hence of the cation, [M+].

Step 2: Write the equation for the hydrolysis of the cation.

Step 3: Use a R.I.C.E. Table to determine the equilibrium concentration of all species.

Step 4: Calculate the value of the cation hydrolysis constant, Kh (acid dissociation constant, Ka) if not given.

Step 5: Calculate the concentration of hydrogen ions using Ka

Step 6: Calculate the pH of the solution.

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Footnotes

(1) Sulfuric acid is a diprotic acid for which the first dissocation is strong, but the second is weaker. The salt of sulfuric acid and a strong base will be a bit higher than 7 (for aqueous solutions at 25°C)

(2) What about the salts of polyprotic acids? This becomes a little more complicated especially if we are considering the pH of the so-called "acid salts" which are the salts produced by the removal of only some of the protons of a polyprotic acid. Examples of "acid salts" are sodium hydrogencarbonate (NaHCO3) and sodium hydrogensulfite (NaHSO3). The anions in "acid salts" are amphiprotic, they can act like an acid or a base. Just like the problem with the salts of weak acid and weak base, we need to decide which is more likely to occur; the acid dissociation of the anion or the base dissociation of the anion.
We will leave this discussion for another tutorial.

(3) H3O+ has the systematic IUPAC name of oxidanium, and the acceptable non-sytematic IUPAC name of oxonium.
In older textbooks you may also see this referred to as the hydronium ion, but this is no longer an acceptable IUPAC name.
Refer to the "Nomenclature of Inorganic Chemistry: IUPAC Recommendations 2005" (Red Book) for information about naming inorganic compounds.